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2016-01-02T13:54:09+05:30

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Question: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60.

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ANSWER:

There are three consecutive integers (same as, natural 
numbers).

let the smallest of the numbers be 
x.

Values of 
x :

1st no. is x
2nd no. is x+1.
3rd no. is x+1+1  =x+2.

Condition: The square of the 2nd no. (middle number) is 60 more than the difference of the squares of other two numbers.

Solution:

(x+1)^{2}=(x+2)^{2}-x^{2}

=>   x^{2}+1= x^{2}+4-x^{2}

=>  
 x^{2}-x^{2}+x^{2}=4-1

=>   (x^{2}-x^{2})+x^{2}=4-1

=>  0+x^{2}=3

=>  x^{2}=3

=>  x= \sqrt{3}


By putting the values of x,

1st no. is x= \sqrt{3} .
2nd no. is x+1= \sqrt{3}+1 .
3rd no. is x+2= \sqrt{3}+2 .

ANSWER: The values of the three consecutive numbers:

1st no. is x= \sqrt{3} .
2nd no. is x+1= \sqrt{3}+1 .
3rd no. is x+2= \sqrt{3}+2 .


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